Question: Let $y=x^4\ln(x)$. Find $\dfrac{dy}{dx}$. Choose 1 answer: Choose 1 answer: (Choice A) A $4x^3+\dfrac1x$ (Choice B) B $4x^3(x^4+\ln(x))$ (Choice C) C $x^3(4\ln(x)+1)$ (Choice D) D $4x^2$
Solution: $x^4\ln(x)$ is the product of two, more basic, expressions: $x^4$ and $\ln(x)$. Therefore, $\dfrac{dy}{dx}$ can be found using the product rule : $\begin{aligned} \dfrac{d}{dx}[u(x)v(x)]&=\dfrac{d}{dx}[u(x)]v(x)+u(x)\dfrac{d}{dx}[v(x)] \\\\ &=u'(x)v(x)+u(x)v'(x) \end{aligned}$ $\begin{aligned} &\phantom{=}\dfrac{dy}{dx} \\\\ &=\dfrac{d}{dx}(x^4\ln(x)) \\\\ &=\dfrac{d}{dx}(x^4)\ln(x)+x^4\dfrac{d}{dx}(\ln(x))&&\gray{\text{The product rule}} \\\\ &=4x^3\cdot \ln(x)+x^4\cdot \dfrac1x&&\gray{\text{Differentiate }x^4\text{ and }\ln(x)} \\\\ &=4x^3\ln(x)+x^3&&\gray{\text{Simplify}} \\\\ &=x^3(4\ln(x)+1)&&\gray{\text{Factor out }x^3} \end{aligned}$ In conclusion, $\dfrac{dy}{dx}=x^3(4\ln(x)+1)$